Elements of Algebra

202 ELEMENTS SECT. IV. The banker must therefore give $\frac{a(b-c)}{b-a}$ pieces of the first kind, and $\frac{b(c-a)}{b-a}$ pieces of the second kind. *Remark*. These two numbers are easily found by the Rule of Three, when it is required to apply the results which we have obtained. Thus, to find the first we say, $b-a:a::b-c:\frac{a(b-c)}{b-a}$; and the second number is found thus; $b-a:b::c-a:\frac{b(c-a)}{b-a}$. It ought to be observed also, that $a$ is less than $b$, and that $c$ is less than $b$; but at the same time greater than $a$, as the nature of the thing requires. 603. *Question* 20. A banker has two kinds of change; 10 pieces of one make a crown, and 20 pieces of the other make a crown; and a person wishes to change a crown into 17 pieces of money: how many of each sort must he have? We have here $a=10$, $b=20$, and $c=17$, which furnishes the following proportions: First, $10:10::3:3$, so that the number of pieces of the first kind is 3. Secondly, $10:20::7:14$, and the number of the second kind is 14. 604. *Question* 21. A father leaves at his death several children, who share his property in the following manner: namely, the first receives a hundred pounds, and the tenth part of the remainder; the second receives two hundred pounds, and the tenth part of the remainder; the third takes three hundred pounds, and the tenth part of what remains; and the fourth takes four hundred pounds, and the tenth part of what then remains; and so on. And it is found that the property has thus been divided equally among all the children. Required how much it was, how many children there were, and how much each received? This question is rather of a singular nature, and therefore deserves particular attention. In order to resolve it more easily, we shall suppose the whole fortune to be $z$ pounds; and since all the children receive the same sum, let the share of each be $x$, by which means the number of children will be expressed by $\frac{z}{x}$. Now, this being laid down, we may proceed to the solution of the question, as follows: